Problem: The lifespans of turtles in a particular zoo are normally distributed. The average turtle lives $83$ years; the standard deviation is $11.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a turtle living less than $117.2$ years.
$83$ $71.6$ $94.4$ $60.2$ $105.8$ $48.8$ $117.2$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $83$ years. We know the standard deviation is $11.4$ years, so one standard deviation below the mean is $71.6$ years and one standard deviation above the mean is $94.4$ years. Two standard deviations below the mean is $60.2$ years and two standard deviations above the mean is $105.8$ years. Three standard deviations below the mean is $48.8$ years and three standard deviations above the mean is $117.2$ years. We are interested in the probability of a turtle living less than $117.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the turtles will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the turtles will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $48.8$ years and the other half $({0.15\%})$ will live longer than $117.2$ years. The probability of a particular turtle living less than $117.2$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.